Question

One quarter sector is cut from a uniform circular disc of radius \(\mathrm{R}\). This sector has mass \(\mathrm{M}\). It is made to rotate about a line perpendicular to its plane and passing through the centre of the original disc. Its moment of inertia about the axis of rotation is... \(\\{\mathrm{A}\\}(1 / 2) \mathrm{MR}^{2}\) \(\\{\mathrm{B}\\}(1 / 4) \mathrm{MR}^{2}\) \(\\{\mathrm{C}\\}(1 / 8) \mathrm{MR}^{2}\) \\{D \(\\} \sqrt{2} \mathrm{MR}^{2}\)

Short answer

The moment of inertia of the quarter sector is \(\cfrac{1}{4} MR^2\), which corresponds to answer choice B.

Step-by-step solution

Identify the variables

We are given the radius of the circular disc, R, and the mass of the quarter sector, M.

Calculate the moment of inertia of the circular disc

We first calculate the moment of inertia of the complete uniform circular disc, which is given by the equation \(I_{disc} = \cfrac{1}{2}MR^2\), where M is the mass and R is the radius of the disc.

Calculate the mass of the original disc

We need to find the mass of the original disc to determine the mass density. Since the quarter sector has mass M, the original disc has a mass of 4M.

Calculate the mass density of the disc

We can find the mass density, ρ, of the original disc by dividing the total mass by the total area. The mass density ρ is given by: \[ρ = \cfrac{4M}{πR^2}\]

Calculate the moment of inertia of the sector

For the quarter sector, we first calculate the moment of inertia for a small mass element, dm, with respect to the axis of rotation. Then, we will integrate over the entire quarter sector. The mass element dm can be written as: \[dm = ρ \cdot dA\] where dA = r dr dθ, and θ is the angle between the outer radius and the quarter sector. So, dm = \(\frac{4M}{πR^2} rdrdθ\) Now, let's plug dm into the moment of inertia equation: \[dI = r^2 dm = r^2 \cdot \frac{4M}{πR^2} rdrdθ\] Next, we integrate dI to find the moment of inertia I of the quarter sector: \[I_{sector} = \int\int r^2 \cdot \frac{4M}{πR^2} rdrdθ\] For the quarter sector, we integrate from r=0 to R and θ=0 to π/2: \[I_{sector} = \cfrac{4M}{πR^2} \int_{0}^{R} r^3 dr \int_{0}^{\frac{π}{2}} dθ\]

Perform the integrations and simplify

Perform the integrations and simplify the expression: \[I_{sector} = \cfrac{4M}{πR^2} \cdot \left[\cfrac{r^4}{4}\right]_{0}^{R} \cdot \left[θ\right]_{0}^{\frac{π}{2}}\] \[I_{sector} = \cfrac{1}{4} MR^2\] The moment of inertia of the quarter sector is (1/4)MR², which corresponds to answer choice B.